Q. 37 L5.0( 1 Vote )

Solve each of the

Answer :

, y = 2 when


Given and



This is a first order linear differential equation of the form



Here, P = –3 cot x and Q = sin 2x


The integrating factor (I.F) of this differential equation is,





We have



[ m log a = log am]




I.F = cosec3x [ elog x = x]


Hence, the solution of the differential equation is,








Recall


ycosec3x = 2(–cosec x) + c


ycosec3x = –2cosec x + c




y = –2sin2x + csin3x


However, when, we have y = 2



2 = –2(1)2 + c(1)3


2 = –2 + c


c = 4


By substituting the value of c in the equation for y, we get


y = –2sin2x + (4)sin3x


y = –2sin2x + 4sin3x


Thus, the solution of the given initial value problem is y = –2sin2x + 4sin3x


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